Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct P {
  int x, y;
};
struct matSpace {
  vector<P> ops;
  int data[10010];
  int rank, len;
  void input(int n) {
    len = n;
    for (int i = 0; i < len; ++i) {
      scanf("%d", &data[i]);
    }
    gauss();
  }
  void gauss() {
    rank = 0;
    ops.clear();
    for (int b = 30; b >= 0; --b) {
      int mainRowID = -1;
      for (int i = rank; i < len; ++i) {
        if ((data[i] >> b) & 1) {
          mainRowID = i;
          break;
        }
      }
      if (mainRowID != -1) {
        if (mainRowID != rank) {
          doSwap(mainRowID, rank);
        }
        for (int j = rank + 1; j < len; ++j) {
          if ((data[j] >> b) & 1) {
            doXor(j, rank);
          }
        }
        rank += 1;
      }
    }
  }
  void doSwap(int x, int y) {
    ops.push_back(P{x, y});
    ops.push_back(P{y, x});
    ops.push_back(P{x, y});
    swap(data[x], data[y]);
  }
  void doXor(int x, int y) {
    ops.push_back(P{x, y});
    data[x] ^= data[y];
  }
  void adjust(const matSpace &G) {
    int maxRank = max(rank, G.rank);
    int colID = 30;
    for (int r = 0; r < rank; ++r) {
      while (((data[r] >> colID) & 1) == 0) {
        colID -= 1;
      }
      for (int i = 0; i < maxRank; ++i) {
        if (((data[i] >> colID) & 1) != ((G.data[i] >> colID) & 1)) {
          doXor(i, r);
        }
      }
    }
  }
  bool isSameSpace(const matSpace &G) {
    int maxRank = max(rank, G.rank);
    for (int r = 0; r < maxRank; ++r) {
      if (data[r] != G.data[r]) {
        return false;
      }
    }
    return true;
  }
  int getOpSize() { return (int)ops.size(); }
  void outputOps(bool rev = false) {
    if (rev) {
      reverse(ops.begin(), ops.end());
    }
    for (const auto &op : ops) {
      printf("%d %d\n", op.x + 1, op.y + 1);
    }
  }
};
matSpace X, Y;
int n;
int main() {
  scanf("%d", &n);
  X.input(n);
  Y.input(n);
  X.adjust(Y);
  if (!X.isSameSpace(Y)) {
    puts("-1");
  } else {
    printf("%d\n", X.getOpSize() + Y.getOpSize());
    X.outputOps();
    Y.outputOps(true);
  }
  return 0;
}